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A rectangular strip of paper edge $AC$ is first folded in half making fold crease through mid-point $M$ and again bent folded along a new line through $C$ adjusted such that corner $A$ falls on divider $MB$. Show that triangle $ABC$ is equilateral.

EDIT1:

Rajendra Kumar makes a full Icosahedron repeated in this way!

Origami

Narasimham
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    By construction of $B$, $BC = AC$. By symmetry wrt line $MB$, $AB = CB$... Am I missing something??? – achille hui Oct 15 '18 at 22:23
  • Nothing, :) except that it's too simple! – Narasimham Oct 16 '18 at 03:48
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    Related: This response which offers the answer "tetrahedron" to the question "Is there a shape that can be wrapped perfectly?"; the wrapping technique involves this folding construction of the equilateral triangle. My comment there notes that U.S. dollar bills are just-about perfect for folding such tetrahedra. – Blue Oct 16 '18 at 09:32
  • Even an irregular/smaller octahedron possible by folding dollar bill into triangles, no? – Narasimham Oct 22 '18 at 00:05

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This is an interesting problem. Thank you for posting it.

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Construct $BD$ perpendicular to one edge of the paper $$ BD = MC = \frac{1}{2}AC = \frac{1}{2}BC$$ It follows that $$\angle BCD = 30^\circ\\ \angle MCB = 60^\circ $$ From the condition that you give, $$AC = BC$$ Therefore $\Delta ACB$ is an equilateral triangle.

Larry
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